The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experimental conditions (“experimental measurement of the stopping performance of a tractor-semitrailer from multiple speeds,” nhtsa, dot hs 811 488, june 2011): 32.1 30.6 31.4 30.4 31.0 31.9 the cited report states that under these conditions, the maximum allowable stopping distance is 30. a normal probability plot validates the assumption that stopping distance is normally distributed. a. does the data suggest that true average stopping distance exceeds this maximum value? test the appropriate hypotheses using a 5 .01.

There is enough evidence at the 1% level of significance to suggest that the mean stopping distance of the truck is greater than 30

Step-by-step explanation:

We will conduct a hypothesis test for the mean.  We will compare the mean stopping distance of the truck to the maximum distance of 30 and see if there is any significant difference.  

We first need to find the mean and standard deviation of the stopping distance of the sample.  See the first attached photo for the calculations of these values..

We get a sample mean of: 31.2333

and a sample standard deviation of: 0.2813

We will use these to conduct a hypothesis test.  See the calculations for the test on the second attached photo

step-by-step explanation:

. your answer will contain x - and y -terms. (1 point) x ( ¿¿ 2 − 7 ) 2 +( y 2 − 9 ) 2 √ ¿ 12. what is the distance from the directrix (the crocodile river) to the point ( x , y )? write this equation. your answer will contain a y -term.

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.